let X=(C01([0,1],R),‖⋅‖1)X=\Big( \mathcal{C}_1^0 \big([0,1], \mathbb{R}\big), \|\cdot \|_1 \Big) be the space of real-valued continuous functions on [0, 1][0, 1] with the norm \|\cdot \|_1\|\cdot \|_1, \| f \|_1:= \int_0^1 |f(t)|\,dt. \| f \|_1:= \int_0^1 |f(t)|\,dt.

Let us also denote the unit sphere in XX by S_X:= \big\{ f\in X: \|f \|_1 =1 \big\}S_X:= \big\{ f\in X: \|f \|_1 =1 \big\}.

Is S_XS_X totally bounded in X X?

I bet it is not, but I can’t manage to prove it. First, while S_XS_X is not compact, XX is not a Banach space, so one can’t use that a subset AA of a metric space YY is compact iff it is complete and totally bounded. (This argument works flawlessly for every infinite-dimensional Banach space, but not here)

I’ve also tried to get a contradiction by taking several equivalences and implications on being totally bounded, and thereafter supposing that S_XS_X is totally bounded, but none of them works here (or at least none of those of which I’m aware of)

Can you provide me with help?

Thanks in advance!

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I forgot to add that no Riesz’ lemma is allowed here (I’m taking a course in Real Analysis, though a bit tough)

– EternalBlood

Oct 21 at 2:49

Work from the definition. A finite number of continuous functions on [0,1][0,1] are uniformly bounded. Construct a function of unit norm that is more than {1 \over 4}{1 \over 4} away from the finite set.

– copper.hat

Oct 21 at 3:01

You can use the completion of XX whose closed unit ball is the closure of the unit ball of XX and hence compact (because the unit ball of XX is totally bounded=precompact).

– Jochen

2 days ago

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1 Answer

1

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Hint: In SS you can find f_1, f_2, \dots f_1, f_2, \dots such that d(f_m,f_n)=2d(f_m,f_n)=2 for m\ne n.m\ne n. Why? Because you can find pairwise disjoint closed intervals I_1,I_2, \dotsI_1,I_2, \dots of positive length contained in [0,1].[0,1].